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Why would five hundred and five be same as one?
What's the teacher's fractional addition algorithm?One hundred and one hatsGrandpa can't even tell time? Another Grandpa Mystery1 = 2? A Non-Math RiddleSimple Math Problem #3$6$ is prime? Another Grandpa Mystery (not made by @DEEM, this time)A Body Part inside another Body PartMy Grandpa's Box #1Can 4 = 6 and 10 = 1000? : Another Grandpa QuestionI walked uphill both ways
$begingroup$
Grandpa was in his crazy math mood again.
"using rot13 and your math knowledge prove to me that
505 = 1"
He said.
Really? Can you?
HINT
Think Trigonometry
riddle mathematics lateral-thinking
edited Mar 20 at 20:48
JonMark Perry
20.6k64099
asked Mar 19 at 13:02
DEEMDEEM
6,516121116
$endgroup$
add a comment |
$begingroup$
Grandpa was in his crazy math mood again.
"using rot13 and your math knowledge prove to me that
505 = 1"
He said.
Really? Can you?
HINT
Think Trigonometry
riddle mathematics lateral-thinking
edited Mar 20 at 20:48
JonMark Perry
20.6k64099
asked Mar 19 at 13:02
DEEMDEEM
6,516121116
$endgroup$
4
$begingroup$
Don't have an answer, but I'll note 2 findings given the hint: 1) 505 is the hypotenuse of a triangle with integer length sides. 2) Rot13-ing the Roman Numerals for the number 105 produces a trigonometric term. Maybe one of these is on the right path and will help someone.
$endgroup$
– TwoBitOperation
Mar 20 at 14:40
3
$begingroup$
@TwoBitOperation for anyone possibly going with route 1: The possible sides of a triangle that work are (100, 495) (217, 456) (303, 404) (336, 377)
$endgroup$
– akozi
Mar 20 at 16:28
add a comment |
$begingroup$
Grandpa was in his crazy math mood again.
"using rot13 and your math knowledge prove to me that
505 = 1"
He said.
Really? Can you?
HINT
Think Trigonometry
riddle mathematics lateral-thinking
edited Mar 20 at 20:48
JonMark Perry
20.6k64099
asked Mar 19 at 13:02
DEEMDEEM
6,516121116
$endgroup$
Grandpa was in his crazy math mood again.
"using rot13 and your math knowledge prove to me that
505 = 1"
He said.
Really? Can you?
HINT
Think Trigonometry
riddle mathematics lateral-thinking
riddle mathematics lateral-thinking
edited Mar 20 at 20:48
JonMark Perry
20.6k64099
asked Mar 19 at 13:02
DEEMDEEM
6,516121116
edited Mar 20 at 20:48
JonMark Perry
20.6k64099
asked Mar 19 at 13:02
DEEMDEEM
6,516121116
edited Mar 20 at 20:48
JonMark Perry
20.6k64099
edited Mar 20 at 20:48
JonMark Perry
20.6k64099
edited Mar 20 at 20:48
JonMark Perry
20.6k64099
20.6k64099
asked Mar 19 at 13:02
DEEMDEEM
6,516121116
asked Mar 19 at 13:02
DEEMDEEM
6,516121116
asked Mar 19 at 13:02
DEEMDEEM
6,516121116
6,516121116
4
$begingroup$
Don't have an answer, but I'll note 2 findings given the hint: 1) 505 is the hypotenuse of a triangle with integer length sides. 2) Rot13-ing the Roman Numerals for the number 105 produces a trigonometric term. Maybe one of these is on the right path and will help someone.
$endgroup$
– TwoBitOperation
Mar 20 at 14:40
3
$begingroup$
@TwoBitOperation for anyone possibly going with route 1: The possible sides of a triangle that work are (100, 495) (217, 456) (303, 404) (336, 377)
$endgroup$
– akozi
Mar 20 at 16:28
add a comment |
4
$begingroup$
Don't have an answer, but I'll note 2 findings given the hint: 1) 505 is the hypotenuse of a triangle with integer length sides. 2) Rot13-ing the Roman Numerals for the number 105 produces a trigonometric term. Maybe one of these is on the right path and will help someone.
$endgroup$
– TwoBitOperation
Mar 20 at 14:40
3
$begingroup$
@TwoBitOperation for anyone possibly going with route 1: The possible sides of a triangle that work are (100, 495) (217, 456) (303, 404) (336, 377)
$endgroup$
– akozi
Mar 20 at 16:28
4
4
$begingroup$
Don't have an answer, but I'll note 2 findings given the hint: 1) 505 is the hypotenuse of a triangle with integer length sides. 2) Rot13-ing the Roman Numerals for the number 105 produces a trigonometric term. Maybe one of these is on the right path and will help someone.
$endgroup$
– TwoBitOperation
Mar 20 at 14:40
$begingroup$
Don't have an answer, but I'll note 2 findings given the hint: 1) 505 is the hypotenuse of a triangle with integer length sides. 2) Rot13-ing the Roman Numerals for the number 105 produces a trigonometric term. Maybe one of these is on the right path and will help someone.
$endgroup$
– TwoBitOperation
Mar 20 at 14:40
3
3
$begingroup$
@TwoBitOperation for anyone possibly going with route 1: The possible sides of a triangle that work are (100, 495) (217, 456) (303, 404) (336, 377)
$endgroup$
– akozi
Mar 20 at 16:28
$begingroup$
@TwoBitOperation for anyone possibly going with route 1: The possible sides of a triangle that work are (100, 495) (217, 456) (303, 404) (336, 377)
$endgroup$
– akozi
Mar 20 at 16:28
add a comment |
9 Answers
9
active
oldest
votes
$begingroup$
Here's one possible solution:
1. $505 = 1$ : Given
2. $DV = 1$ : By Romanizing left
3. $V = 1/D$
4. $CV = C/D$ : Multiply through by 100
5. $(PI)= C/D$ : By Rot(13)ing left
6. $Dπ = C$
By the circumference formula for a circle (C= πD),the left and right are equivalent
Q.E.D.
answered Mar 20 at 20:19
TwoBitOperationTwoBitOperation
8,20911465
$endgroup$
11
$begingroup$
Wow. If that's the intended solution I'll be quite surprised. Regardless, that's an amazingly convoluted way to answer this, and I love it. :)
$endgroup$
– Rubio♦
Mar 20 at 20:51
5
$begingroup$
I'm... just as surprised as everyone else here.
$endgroup$
– TwoBitOperation
Mar 20 at 21:44
8
$begingroup$
@DEEM What does this have to do with trigonometry?
$endgroup$
– noedne
Mar 20 at 21:53
4
$begingroup$
wouldn't step 5 require you to to do both sides, yieldingP/Q?
$endgroup$
– corsiKa
Mar 21 at 0:49
8
$begingroup$
Step 2 to 3 doesn't make sense, as Roman numerals are additive, not multiplicative.
$endgroup$
– Herb Wolfe
Mar 21 at 3:19
|
show 4 more comments
$begingroup$
Reasoning
In Roman numerals 505 is DV. If we use rot13 on these two characters, we get QI.
Qi is the circulating life force whose existence and properties are the basis of much Chinese philosophy and medicine. It allows us to say that we are one with the universe. In this way, 505=1
answered Mar 19 at 13:08
hexominohexomino
46.1k4140219
$endgroup$
$begingroup$
Lot different than my answer
$endgroup$
– DEEM
Mar 19 at 13:10
1
$begingroup$
@DEEM Can you tell us a way in which your answer differs from this answer?
$endgroup$
– Tanner Swett
Mar 19 at 18:48
1
$begingroup$
So there is the rot13 part and there is the math part. Both are needed for my answer
$endgroup$
– DEEM
Mar 20 at 1:36
add a comment |
$begingroup$
By using
hexomino's idea of ROT13 on Roman numerals
we can obtain
$text{ROT13}(505)=text{ROT13}(V^4-V^3+V)=I^4-I^3+I=1$.
answered Mar 19 at 15:27
noednenoedne
8,67512365
$endgroup$
$begingroup$
Could you (or one of the 22 upvoters) explain the last step to me?
$endgroup$
– user1717828
Mar 21 at 1:35
$begingroup$
@user1717828 $I$ is the Roman numeral for $1$, so $I^4-I^3+I=1^4-1^3+1=1-1+1=1$.
$endgroup$
– noedne
Mar 21 at 2:03
$begingroup$
Ahh!! I did the complex math out like three times before submitting to Google to confirm I wasn't crazy. No idea what I was thinking. Thanks!
$endgroup$
– user1717828
Mar 21 at 10:46
add a comment |
$begingroup$
The rot13 translation of "one" is "bar". The only connection between "bar" and 505 that I could find is this very obscure definition given in urban dictionary (which was entry #4 and has more downvotes than upvotes), which says that it's slang for getting a drink at a bar with friends after work. Perhaps it's a really old outdated slang term, which is why your grandpa used it.
answered Mar 19 at 16:38
BridgeburnersBridgeburners
1612
$endgroup$
$begingroup$
Beer O'Clock, Can't be any different to 4:20 or the Masonic arguments pertaining to yardarms or the time of day, It's beer O'Clock somewhere in the world. In the olden days 505 meant OOO. We left at 5pm.
$endgroup$
– mckenzm
Mar 21 at 5:42
add a comment |
$begingroup$
Elaborating on @hexomino answer 505 -> DV.
Is it possible that:
I want to DownVote you 505 times which was odd number resulting 1 DV.
var num=505; isOdd(num); function isOdd(num) { return num % 2;}
I'm not downvoting you.
answered Mar 20 at 10:24
MukyuuMukyuu
5011113
$endgroup$
add a comment |
$begingroup$
Use the digital radix: 5 + 0 + 5 = 10, 1 + 0 = 1
answered Mar 19 at 15:11
LurkerLurker
111
$endgroup$
11
$begingroup$
And where is Rot13 used?
$endgroup$
– Chronocidal
Mar 19 at 15:58
add a comment |
$begingroup$
Working out possible solutions using rot13 and trig:
I feel gross and like I'm doing something wrong doing this but:
505 = 5 * 101 => V * CI
Then you:
rot13(V * CI) = I * PI => i * pi (getting pi honestly is the only reason I think this might be the right direction.
Here's where the rot13(gevt) comes in I guess? Not sure where to proceed from there.
answered Mar 20 at 20:06
Nick VithaNick Vitha
1055
$endgroup$
1
$begingroup$
rot13 of I is not I.
$endgroup$
– Rubio♦
Mar 20 at 20:53
1
$begingroup$
Well I'm an idiot.
$endgroup$
– Nick Vitha
Mar 20 at 21:05
add a comment |
$begingroup$
Tongue in cheek answer - technically works anyhow ;-)
050/5 rdhnyf 6/5 gurersber 050 rdhnyf 6. "Cebirq" hfvat zngu xabjyrqtr naq ebg68, nf erdhrfgrq ;-)
Rot13 a_zA-Z0-9
answered Mar 20 at 22:39
StilezStilez
1,224211
$endgroup$
add a comment |
$begingroup$
If you use a "fivethousandfiftydecimal" base then 505 in this base is equal to 1 in decimal base.
Just like 12 in octal base is equal to 10 in decimal base.
answered Mar 21 at 11:07
BaudoisBaudois
1
$endgroup$
$begingroup$
Welcome to Puzzling.SE! I like the lateral thinking in this answer, but I'm afraid it doesn't quite answer the question, since it doesn't use rot13.
$endgroup$
– Brandon_J
Mar 21 at 13:27
add a comment |
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9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's one possible solution:
1. $505 = 1$ : Given
2. $DV = 1$ : By Romanizing left
3. $V = 1/D$
4. $CV = C/D$ : Multiply through by 100
5. $(PI)= C/D$ : By Rot(13)ing left
6. $Dπ = C$
By the circumference formula for a circle (C= πD),the left and right are equivalent
Q.E.D.
answered Mar 20 at 20:19
TwoBitOperationTwoBitOperation
8,20911465
$endgroup$
11
$begingroup$
Wow. If that's the intended solution I'll be quite surprised. Regardless, that's an amazingly convoluted way to answer this, and I love it. :)
$endgroup$
– Rubio♦
Mar 20 at 20:51
5
$begingroup$
I'm... just as surprised as everyone else here.
$endgroup$
– TwoBitOperation
Mar 20 at 21:44
8
$begingroup$
@DEEM What does this have to do with trigonometry?
$endgroup$
– noedne
Mar 20 at 21:53
4
$begingroup$
wouldn't step 5 require you to to do both sides, yieldingP/Q?
$endgroup$
– corsiKa
Mar 21 at 0:49
8
$begingroup$
Step 2 to 3 doesn't make sense, as Roman numerals are additive, not multiplicative.
$endgroup$
– Herb Wolfe
Mar 21 at 3:19
|
show 4 more comments
$begingroup$
Here's one possible solution:
1. $505 = 1$ : Given
2. $DV = 1$ : By Romanizing left
3. $V = 1/D$
4. $CV = C/D$ : Multiply through by 100
5. $(PI)= C/D$ : By Rot(13)ing left
6. $Dπ = C$
By the circumference formula for a circle (C= πD),the left and right are equivalent
Q.E.D.
answered Mar 20 at 20:19
TwoBitOperationTwoBitOperation
8,20911465
$endgroup$
11
$begingroup$
Wow. If that's the intended solution I'll be quite surprised. Regardless, that's an amazingly convoluted way to answer this, and I love it. :)
$endgroup$
– Rubio♦
Mar 20 at 20:51
5
$begingroup$
I'm... just as surprised as everyone else here.
$endgroup$
– TwoBitOperation
Mar 20 at 21:44
8
$begingroup$
@DEEM What does this have to do with trigonometry?
$endgroup$
– noedne
Mar 20 at 21:53
4
$begingroup$
wouldn't step 5 require you to to do both sides, yieldingP/Q?
$endgroup$
– corsiKa
Mar 21 at 0:49
8
$begingroup$
Step 2 to 3 doesn't make sense, as Roman numerals are additive, not multiplicative.
$endgroup$
– Herb Wolfe
Mar 21 at 3:19
|
show 4 more comments
$begingroup$
Here's one possible solution:
1. $505 = 1$ : Given
2. $DV = 1$ : By Romanizing left
3. $V = 1/D$
4. $CV = C/D$ : Multiply through by 100
5. $(PI)= C/D$ : By Rot(13)ing left
6. $Dπ = C$
By the circumference formula for a circle (C= πD),the left and right are equivalent
Q.E.D.
answered Mar 20 at 20:19
TwoBitOperationTwoBitOperation
8,20911465
$endgroup$
Here's one possible solution:
1. $505 = 1$ : Given
2. $DV = 1$ : By Romanizing left
3. $V = 1/D$
4. $CV = C/D$ : Multiply through by 100
5. $(PI)= C/D$ : By Rot(13)ing left
6. $Dπ = C$
By the circumference formula for a circle (C= πD),the left and right are equivalent
Q.E.D.
answered Mar 20 at 20:19
TwoBitOperationTwoBitOperation
8,20911465
edited Mar 20 at 20:32
answered Mar 20 at 20:19
TwoBitOperationTwoBitOperation
8,20911465
answered Mar 20 at 20:19
TwoBitOperationTwoBitOperation
8,20911465
answered Mar 20 at 20:19
TwoBitOperationTwoBitOperation
8,20911465
8,20911465
11
$begingroup$
Wow. If that's the intended solution I'll be quite surprised. Regardless, that's an amazingly convoluted way to answer this, and I love it. :)
$endgroup$
– Rubio♦
Mar 20 at 20:51
5
$begingroup$
I'm... just as surprised as everyone else here.
$endgroup$
– TwoBitOperation
Mar 20 at 21:44
8
$begingroup$
@DEEM What does this have to do with trigonometry?
$endgroup$
– noedne
Mar 20 at 21:53
4
$begingroup$
wouldn't step 5 require you to to do both sides, yieldingP/Q?
$endgroup$
– corsiKa
Mar 21 at 0:49
8
$begingroup$
Step 2 to 3 doesn't make sense, as Roman numerals are additive, not multiplicative.
$endgroup$
– Herb Wolfe
Mar 21 at 3:19
|
show 4 more comments
11
$begingroup$
Wow. If that's the intended solution I'll be quite surprised. Regardless, that's an amazingly convoluted way to answer this, and I love it. :)
$endgroup$
– Rubio♦
Mar 20 at 20:51
5
$begingroup$
I'm... just as surprised as everyone else here.
$endgroup$
– TwoBitOperation
Mar 20 at 21:44
8
$begingroup$
@DEEM What does this have to do with trigonometry?
$endgroup$
– noedne
Mar 20 at 21:53
4
$begingroup$
wouldn't step 5 require you to to do both sides, yieldingP/Q?
$endgroup$
– corsiKa
Mar 21 at 0:49
8
$begingroup$
Step 2 to 3 doesn't make sense, as Roman numerals are additive, not multiplicative.
$endgroup$
– Herb Wolfe
Mar 21 at 3:19
11
11
$begingroup$
Wow. If that's the intended solution I'll be quite surprised. Regardless, that's an amazingly convoluted way to answer this, and I love it. :)
$endgroup$
– Rubio♦
Mar 20 at 20:51
$begingroup$
Wow. If that's the intended solution I'll be quite surprised. Regardless, that's an amazingly convoluted way to answer this, and I love it. :)
$endgroup$
– Rubio♦
Mar 20 at 20:51
5
5
$begingroup$
I'm... just as surprised as everyone else here.
$endgroup$
– TwoBitOperation
Mar 20 at 21:44
$begingroup$
I'm... just as surprised as everyone else here.
$endgroup$
– TwoBitOperation
Mar 20 at 21:44
8
8
$begingroup$
@DEEM What does this have to do with trigonometry?
$endgroup$
– noedne
Mar 20 at 21:53
$begingroup$
@DEEM What does this have to do with trigonometry?
$endgroup$
– noedne
Mar 20 at 21:53
4
4
$begingroup$
wouldn't step 5 require you to to do both sides, yielding
P/Q?$endgroup$
– corsiKa
Mar 21 at 0:49
$begingroup$
wouldn't step 5 require you to to do both sides, yielding
P/Q?$endgroup$
– corsiKa
Mar 21 at 0:49
8
8
$begingroup$
Step 2 to 3 doesn't make sense, as Roman numerals are additive, not multiplicative.
$endgroup$
– Herb Wolfe
Mar 21 at 3:19
$begingroup$
Step 2 to 3 doesn't make sense, as Roman numerals are additive, not multiplicative.
$endgroup$
– Herb Wolfe
Mar 21 at 3:19
|
show 4 more comments
$begingroup$
Reasoning
In Roman numerals 505 is DV. If we use rot13 on these two characters, we get QI.
Qi is the circulating life force whose existence and properties are the basis of much Chinese philosophy and medicine. It allows us to say that we are one with the universe. In this way, 505=1
answered Mar 19 at 13:08
hexominohexomino
46.1k4140219
$endgroup$
$begingroup$
Lot different than my answer
$endgroup$
– DEEM
Mar 19 at 13:10
1
$begingroup$
@DEEM Can you tell us a way in which your answer differs from this answer?
$endgroup$
– Tanner Swett
Mar 19 at 18:48
1
$begingroup$
So there is the rot13 part and there is the math part. Both are needed for my answer
$endgroup$
– DEEM
Mar 20 at 1:36
add a comment |
$begingroup$
Reasoning
In Roman numerals 505 is DV. If we use rot13 on these two characters, we get QI.
Qi is the circulating life force whose existence and properties are the basis of much Chinese philosophy and medicine. It allows us to say that we are one with the universe. In this way, 505=1
answered Mar 19 at 13:08
hexominohexomino
46.1k4140219
$endgroup$
$begingroup$
Lot different than my answer
$endgroup$
– DEEM
Mar 19 at 13:10
1
$begingroup$
@DEEM Can you tell us a way in which your answer differs from this answer?
$endgroup$
– Tanner Swett
Mar 19 at 18:48
1
$begingroup$
So there is the rot13 part and there is the math part. Both are needed for my answer
$endgroup$
– DEEM
Mar 20 at 1:36
add a comment |
$begingroup$
Reasoning
In Roman numerals 505 is DV. If we use rot13 on these two characters, we get QI.
Qi is the circulating life force whose existence and properties are the basis of much Chinese philosophy and medicine. It allows us to say that we are one with the universe. In this way, 505=1
answered Mar 19 at 13:08
hexominohexomino
46.1k4140219
$endgroup$
Reasoning
In Roman numerals 505 is DV. If we use rot13 on these two characters, we get QI.
Qi is the circulating life force whose existence and properties are the basis of much Chinese philosophy and medicine. It allows us to say that we are one with the universe. In this way, 505=1
answered Mar 19 at 13:08
hexominohexomino
46.1k4140219
answered Mar 19 at 13:08
hexominohexomino
46.1k4140219
answered Mar 19 at 13:08
hexominohexomino
46.1k4140219
answered Mar 19 at 13:08
hexominohexomino
46.1k4140219
46.1k4140219
$begingroup$
Lot different than my answer
$endgroup$
– DEEM
Mar 19 at 13:10
1
$begingroup$
@DEEM Can you tell us a way in which your answer differs from this answer?
$endgroup$
– Tanner Swett
Mar 19 at 18:48
1
$begingroup$
So there is the rot13 part and there is the math part. Both are needed for my answer
$endgroup$
– DEEM
Mar 20 at 1:36
add a comment |
$begingroup$
Lot different than my answer
$endgroup$
– DEEM
Mar 19 at 13:10
1
$begingroup$
@DEEM Can you tell us a way in which your answer differs from this answer?
$endgroup$
– Tanner Swett
Mar 19 at 18:48
1
$begingroup$
So there is the rot13 part and there is the math part. Both are needed for my answer
$endgroup$
– DEEM
Mar 20 at 1:36
$begingroup$
Lot different than my answer
$endgroup$
– DEEM
Mar 19 at 13:10
$begingroup$
Lot different than my answer
$endgroup$
– DEEM
Mar 19 at 13:10
1
1
$begingroup$
@DEEM Can you tell us a way in which your answer differs from this answer?
$endgroup$
– Tanner Swett
Mar 19 at 18:48
$begingroup$
@DEEM Can you tell us a way in which your answer differs from this answer?
$endgroup$
– Tanner Swett
Mar 19 at 18:48
1
1
$begingroup$
So there is the rot13 part and there is the math part. Both are needed for my answer
$endgroup$
– DEEM
Mar 20 at 1:36
$begingroup$
So there is the rot13 part and there is the math part. Both are needed for my answer
$endgroup$
– DEEM
Mar 20 at 1:36
add a comment |
$begingroup$
By using
hexomino's idea of ROT13 on Roman numerals
we can obtain
$text{ROT13}(505)=text{ROT13}(V^4-V^3+V)=I^4-I^3+I=1$.
answered Mar 19 at 15:27
noednenoedne
8,67512365
$endgroup$
$begingroup$
Could you (or one of the 22 upvoters) explain the last step to me?
$endgroup$
– user1717828
Mar 21 at 1:35
$begingroup$
@user1717828 $I$ is the Roman numeral for $1$, so $I^4-I^3+I=1^4-1^3+1=1-1+1=1$.
$endgroup$
– noedne
Mar 21 at 2:03
$begingroup$
Ahh!! I did the complex math out like three times before submitting to Google to confirm I wasn't crazy. No idea what I was thinking. Thanks!
$endgroup$
– user1717828
Mar 21 at 10:46
add a comment |
$begingroup$
By using
hexomino's idea of ROT13 on Roman numerals
we can obtain
$text{ROT13}(505)=text{ROT13}(V^4-V^3+V)=I^4-I^3+I=1$.
answered Mar 19 at 15:27
noednenoedne
8,67512365
$endgroup$
$begingroup$
Could you (or one of the 22 upvoters) explain the last step to me?
$endgroup$
– user1717828
Mar 21 at 1:35
$begingroup$
@user1717828 $I$ is the Roman numeral for $1$, so $I^4-I^3+I=1^4-1^3+1=1-1+1=1$.
$endgroup$
– noedne
Mar 21 at 2:03
$begingroup$
Ahh!! I did the complex math out like three times before submitting to Google to confirm I wasn't crazy. No idea what I was thinking. Thanks!
$endgroup$
– user1717828
Mar 21 at 10:46
add a comment |
$begingroup$
By using
hexomino's idea of ROT13 on Roman numerals
we can obtain
$text{ROT13}(505)=text{ROT13}(V^4-V^3+V)=I^4-I^3+I=1$.
answered Mar 19 at 15:27
noednenoedne
8,67512365
$endgroup$
By using
hexomino's idea of ROT13 on Roman numerals
we can obtain
$text{ROT13}(505)=text{ROT13}(V^4-V^3+V)=I^4-I^3+I=1$.
answered Mar 19 at 15:27
noednenoedne
8,67512365
answered Mar 19 at 15:27
noednenoedne
8,67512365
answered Mar 19 at 15:27
noednenoedne
8,67512365
answered Mar 19 at 15:27
noednenoedne
8,67512365
8,67512365
$begingroup$
Could you (or one of the 22 upvoters) explain the last step to me?
$endgroup$
– user1717828
Mar 21 at 1:35
$begingroup$
@user1717828 $I$ is the Roman numeral for $1$, so $I^4-I^3+I=1^4-1^3+1=1-1+1=1$.
$endgroup$
– noedne
Mar 21 at 2:03
$begingroup$
Ahh!! I did the complex math out like three times before submitting to Google to confirm I wasn't crazy. No idea what I was thinking. Thanks!
$endgroup$
– user1717828
Mar 21 at 10:46
add a comment |
$begingroup$
Could you (or one of the 22 upvoters) explain the last step to me?
$endgroup$
– user1717828
Mar 21 at 1:35
$begingroup$
@user1717828 $I$ is the Roman numeral for $1$, so $I^4-I^3+I=1^4-1^3+1=1-1+1=1$.
$endgroup$
– noedne
Mar 21 at 2:03
$begingroup$
Ahh!! I did the complex math out like three times before submitting to Google to confirm I wasn't crazy. No idea what I was thinking. Thanks!
$endgroup$
– user1717828
Mar 21 at 10:46
$begingroup$
Could you (or one of the 22 upvoters) explain the last step to me?
$endgroup$
– user1717828
Mar 21 at 1:35
$begingroup$
Could you (or one of the 22 upvoters) explain the last step to me?
$endgroup$
– user1717828
Mar 21 at 1:35
$begingroup$
@user1717828 $I$ is the Roman numeral for $1$, so $I^4-I^3+I=1^4-1^3+1=1-1+1=1$.
$endgroup$
– noedne
Mar 21 at 2:03
$begingroup$
@user1717828 $I$ is the Roman numeral for $1$, so $I^4-I^3+I=1^4-1^3+1=1-1+1=1$.
$endgroup$
– noedne
Mar 21 at 2:03
$begingroup$
Ahh!! I did the complex math out like three times before submitting to Google to confirm I wasn't crazy. No idea what I was thinking. Thanks!
$endgroup$
– user1717828
Mar 21 at 10:46
$begingroup$
Ahh!! I did the complex math out like three times before submitting to Google to confirm I wasn't crazy. No idea what I was thinking. Thanks!
$endgroup$
– user1717828
Mar 21 at 10:46
add a comment |
$begingroup$
The rot13 translation of "one" is "bar". The only connection between "bar" and 505 that I could find is this very obscure definition given in urban dictionary (which was entry #4 and has more downvotes than upvotes), which says that it's slang for getting a drink at a bar with friends after work. Perhaps it's a really old outdated slang term, which is why your grandpa used it.
answered Mar 19 at 16:38
BridgeburnersBridgeburners
1612
$endgroup$
$begingroup$
Beer O'Clock, Can't be any different to 4:20 or the Masonic arguments pertaining to yardarms or the time of day, It's beer O'Clock somewhere in the world. In the olden days 505 meant OOO. We left at 5pm.
$endgroup$
– mckenzm
Mar 21 at 5:42
add a comment |
$begingroup$
The rot13 translation of "one" is "bar". The only connection between "bar" and 505 that I could find is this very obscure definition given in urban dictionary (which was entry #4 and has more downvotes than upvotes), which says that it's slang for getting a drink at a bar with friends after work. Perhaps it's a really old outdated slang term, which is why your grandpa used it.
answered Mar 19 at 16:38
BridgeburnersBridgeburners
1612
$endgroup$
$begingroup$
Beer O'Clock, Can't be any different to 4:20 or the Masonic arguments pertaining to yardarms or the time of day, It's beer O'Clock somewhere in the world. In the olden days 505 meant OOO. We left at 5pm.
$endgroup$
– mckenzm
Mar 21 at 5:42
add a comment |
$begingroup$
The rot13 translation of "one" is "bar". The only connection between "bar" and 505 that I could find is this very obscure definition given in urban dictionary (which was entry #4 and has more downvotes than upvotes), which says that it's slang for getting a drink at a bar with friends after work. Perhaps it's a really old outdated slang term, which is why your grandpa used it.
answered Mar 19 at 16:38
BridgeburnersBridgeburners
1612
$endgroup$
The rot13 translation of "one" is "bar". The only connection between "bar" and 505 that I could find is this very obscure definition given in urban dictionary (which was entry #4 and has more downvotes than upvotes), which says that it's slang for getting a drink at a bar with friends after work. Perhaps it's a really old outdated slang term, which is why your grandpa used it.
answered Mar 19 at 16:38
BridgeburnersBridgeburners
1612
answered Mar 19 at 16:38
BridgeburnersBridgeburners
1612
answered Mar 19 at 16:38
BridgeburnersBridgeburners
1612
answered Mar 19 at 16:38
BridgeburnersBridgeburners
1612
1612
$begingroup$
Beer O'Clock, Can't be any different to 4:20 or the Masonic arguments pertaining to yardarms or the time of day, It's beer O'Clock somewhere in the world. In the olden days 505 meant OOO. We left at 5pm.
$endgroup$
– mckenzm
Mar 21 at 5:42
add a comment |
$begingroup$
Beer O'Clock, Can't be any different to 4:20 or the Masonic arguments pertaining to yardarms or the time of day, It's beer O'Clock somewhere in the world. In the olden days 505 meant OOO. We left at 5pm.
$endgroup$
– mckenzm
Mar 21 at 5:42
$begingroup$
Beer O'Clock, Can't be any different to 4:20 or the Masonic arguments pertaining to yardarms or the time of day, It's beer O'Clock somewhere in the world. In the olden days 505 meant OOO. We left at 5pm.
$endgroup$
– mckenzm
Mar 21 at 5:42
$begingroup$
Beer O'Clock, Can't be any different to 4:20 or the Masonic arguments pertaining to yardarms or the time of day, It's beer O'Clock somewhere in the world. In the olden days 505 meant OOO. We left at 5pm.
$endgroup$
– mckenzm
Mar 21 at 5:42
add a comment |
$begingroup$
Elaborating on @hexomino answer 505 -> DV.
Is it possible that:
I want to DownVote you 505 times which was odd number resulting 1 DV.
var num=505; isOdd(num); function isOdd(num) { return num % 2;}
I'm not downvoting you.
answered Mar 20 at 10:24
MukyuuMukyuu
5011113
$endgroup$
add a comment |
$begingroup$
Elaborating on @hexomino answer 505 -> DV.
Is it possible that:
I want to DownVote you 505 times which was odd number resulting 1 DV.
var num=505; isOdd(num); function isOdd(num) { return num % 2;}
I'm not downvoting you.
answered Mar 20 at 10:24
MukyuuMukyuu
5011113
$endgroup$
add a comment |
$begingroup$
Elaborating on @hexomino answer 505 -> DV.
Is it possible that:
I want to DownVote you 505 times which was odd number resulting 1 DV.
var num=505; isOdd(num); function isOdd(num) { return num % 2;}
I'm not downvoting you.
answered Mar 20 at 10:24
MukyuuMukyuu
5011113
$endgroup$
Elaborating on @hexomino answer 505 -> DV.
Is it possible that:
I want to DownVote you 505 times which was odd number resulting 1 DV.
var num=505; isOdd(num); function isOdd(num) { return num % 2;}
I'm not downvoting you.
answered Mar 20 at 10:24
MukyuuMukyuu
5011113
answered Mar 20 at 10:24
MukyuuMukyuu
5011113
answered Mar 20 at 10:24
MukyuuMukyuu
5011113
answered Mar 20 at 10:24
MukyuuMukyuu
5011113
5011113
add a comment |
add a comment |
$begingroup$
Use the digital radix: 5 + 0 + 5 = 10, 1 + 0 = 1
answered Mar 19 at 15:11
LurkerLurker
111
$endgroup$
11
$begingroup$
And where is Rot13 used?
$endgroup$
– Chronocidal
Mar 19 at 15:58
add a comment |
$begingroup$
Use the digital radix: 5 + 0 + 5 = 10, 1 + 0 = 1
answered Mar 19 at 15:11
LurkerLurker
111
$endgroup$
11
$begingroup$
And where is Rot13 used?
$endgroup$
– Chronocidal
Mar 19 at 15:58
add a comment |
$begingroup$
Use the digital radix: 5 + 0 + 5 = 10, 1 + 0 = 1
answered Mar 19 at 15:11
LurkerLurker
111
$endgroup$
Use the digital radix: 5 + 0 + 5 = 10, 1 + 0 = 1
answered Mar 19 at 15:11
LurkerLurker
111
answered Mar 19 at 15:11
LurkerLurker
111
answered Mar 19 at 15:11
LurkerLurker
111
answered Mar 19 at 15:11
LurkerLurker
111
111
11
$begingroup$
And where is Rot13 used?
$endgroup$
– Chronocidal
Mar 19 at 15:58
add a comment |
11
$begingroup$
And where is Rot13 used?
$endgroup$
– Chronocidal
Mar 19 at 15:58
11
11
$begingroup$
And where is Rot13 used?
$endgroup$
– Chronocidal
Mar 19 at 15:58
$begingroup$
And where is Rot13 used?
$endgroup$
– Chronocidal
Mar 19 at 15:58
add a comment |
$begingroup$
Working out possible solutions using rot13 and trig:
I feel gross and like I'm doing something wrong doing this but:
505 = 5 * 101 => V * CI
Then you:
rot13(V * CI) = I * PI => i * pi (getting pi honestly is the only reason I think this might be the right direction.
Here's where the rot13(gevt) comes in I guess? Not sure where to proceed from there.
answered Mar 20 at 20:06
Nick VithaNick Vitha
1055
$endgroup$
1
$begingroup$
rot13 of I is not I.
$endgroup$
– Rubio♦
Mar 20 at 20:53
1
$begingroup$
Well I'm an idiot.
$endgroup$
– Nick Vitha
Mar 20 at 21:05
add a comment |
$begingroup$
Working out possible solutions using rot13 and trig:
I feel gross and like I'm doing something wrong doing this but:
505 = 5 * 101 => V * CI
Then you:
rot13(V * CI) = I * PI => i * pi (getting pi honestly is the only reason I think this might be the right direction.
Here's where the rot13(gevt) comes in I guess? Not sure where to proceed from there.
answered Mar 20 at 20:06
Nick VithaNick Vitha
1055
$endgroup$
1
$begingroup$
rot13 of I is not I.
$endgroup$
– Rubio♦
Mar 20 at 20:53
1
$begingroup$
Well I'm an idiot.
$endgroup$
– Nick Vitha
Mar 20 at 21:05
add a comment |
$begingroup$
Working out possible solutions using rot13 and trig:
I feel gross and like I'm doing something wrong doing this but:
505 = 5 * 101 => V * CI
Then you:
rot13(V * CI) = I * PI => i * pi (getting pi honestly is the only reason I think this might be the right direction.
Here's where the rot13(gevt) comes in I guess? Not sure where to proceed from there.
answered Mar 20 at 20:06
Nick VithaNick Vitha
1055
$endgroup$
Working out possible solutions using rot13 and trig:
I feel gross and like I'm doing something wrong doing this but:
505 = 5 * 101 => V * CI
Then you:
rot13(V * CI) = I * PI => i * pi (getting pi honestly is the only reason I think this might be the right direction.
Here's where the rot13(gevt) comes in I guess? Not sure where to proceed from there.
answered Mar 20 at 20:06
Nick VithaNick Vitha
1055
answered Mar 20 at 20:06
Nick VithaNick Vitha
1055
answered Mar 20 at 20:06
Nick VithaNick Vitha
1055
answered Mar 20 at 20:06
Nick VithaNick Vitha
1055
1055
1
$begingroup$
rot13 of I is not I.
$endgroup$
– Rubio♦
Mar 20 at 20:53
1
$begingroup$
Well I'm an idiot.
$endgroup$
– Nick Vitha
Mar 20 at 21:05
add a comment |
1
$begingroup$
rot13 of I is not I.
$endgroup$
– Rubio♦
Mar 20 at 20:53
1
$begingroup$
Well I'm an idiot.
$endgroup$
– Nick Vitha
Mar 20 at 21:05
1
1
$begingroup$
rot13 of I is not I.
$endgroup$
– Rubio♦
Mar 20 at 20:53
$begingroup$
rot13 of I is not I.
$endgroup$
– Rubio♦
Mar 20 at 20:53
1
1
$begingroup$
Well I'm an idiot.
$endgroup$
– Nick Vitha
Mar 20 at 21:05
$begingroup$
Well I'm an idiot.
$endgroup$
– Nick Vitha
Mar 20 at 21:05
add a comment |
$begingroup$
Tongue in cheek answer - technically works anyhow ;-)
050/5 rdhnyf 6/5 gurersber 050 rdhnyf 6. "Cebirq" hfvat zngu xabjyrqtr naq ebg68, nf erdhrfgrq ;-)
Rot13 a_zA-Z0-9
answered Mar 20 at 22:39
StilezStilez
1,224211
$endgroup$
add a comment |
$begingroup$
Tongue in cheek answer - technically works anyhow ;-)
050/5 rdhnyf 6/5 gurersber 050 rdhnyf 6. "Cebirq" hfvat zngu xabjyrqtr naq ebg68, nf erdhrfgrq ;-)
Rot13 a_zA-Z0-9
answered Mar 20 at 22:39
StilezStilez
1,224211
$endgroup$
add a comment |
$begingroup$
Tongue in cheek answer - technically works anyhow ;-)
050/5 rdhnyf 6/5 gurersber 050 rdhnyf 6. "Cebirq" hfvat zngu xabjyrqtr naq ebg68, nf erdhrfgrq ;-)
Rot13 a_zA-Z0-9
answered Mar 20 at 22:39
StilezStilez
1,224211
$endgroup$
Tongue in cheek answer - technically works anyhow ;-)
050/5 rdhnyf 6/5 gurersber 050 rdhnyf 6. "Cebirq" hfvat zngu xabjyrqtr naq ebg68, nf erdhrfgrq ;-)
Rot13 a_zA-Z0-9
answered Mar 20 at 22:39
StilezStilez
1,224211
answered Mar 20 at 22:39
StilezStilez
1,224211
answered Mar 20 at 22:39
StilezStilez
1,224211
answered Mar 20 at 22:39
StilezStilez
1,224211
1,224211
add a comment |
add a comment |
$begingroup$
If you use a "fivethousandfiftydecimal" base then 505 in this base is equal to 1 in decimal base.
Just like 12 in octal base is equal to 10 in decimal base.
answered Mar 21 at 11:07
BaudoisBaudois
1
$endgroup$
$begingroup$
Welcome to Puzzling.SE! I like the lateral thinking in this answer, but I'm afraid it doesn't quite answer the question, since it doesn't use rot13.
$endgroup$
– Brandon_J
Mar 21 at 13:27
add a comment |
$begingroup$
If you use a "fivethousandfiftydecimal" base then 505 in this base is equal to 1 in decimal base.
Just like 12 in octal base is equal to 10 in decimal base.
answered Mar 21 at 11:07
BaudoisBaudois
1
$endgroup$
$begingroup$
Welcome to Puzzling.SE! I like the lateral thinking in this answer, but I'm afraid it doesn't quite answer the question, since it doesn't use rot13.
$endgroup$
– Brandon_J
Mar 21 at 13:27
add a comment |
$begingroup$
If you use a "fivethousandfiftydecimal" base then 505 in this base is equal to 1 in decimal base.
Just like 12 in octal base is equal to 10 in decimal base.
answered Mar 21 at 11:07
BaudoisBaudois
1
$endgroup$
If you use a "fivethousandfiftydecimal" base then 505 in this base is equal to 1 in decimal base.
Just like 12 in octal base is equal to 10 in decimal base.
answered Mar 21 at 11:07
BaudoisBaudois
1
answered Mar 21 at 11:07
BaudoisBaudois
1
answered Mar 21 at 11:07
BaudoisBaudois
1
answered Mar 21 at 11:07
BaudoisBaudois
1
1
$begingroup$
Welcome to Puzzling.SE! I like the lateral thinking in this answer, but I'm afraid it doesn't quite answer the question, since it doesn't use rot13.
$endgroup$
– Brandon_J
Mar 21 at 13:27
add a comment |
$begingroup$
Welcome to Puzzling.SE! I like the lateral thinking in this answer, but I'm afraid it doesn't quite answer the question, since it doesn't use rot13.
$endgroup$
– Brandon_J
Mar 21 at 13:27
$begingroup$
Welcome to Puzzling.SE! I like the lateral thinking in this answer, but I'm afraid it doesn't quite answer the question, since it doesn't use rot13.
$endgroup$
– Brandon_J
Mar 21 at 13:27
$begingroup$
Welcome to Puzzling.SE! I like the lateral thinking in this answer, but I'm afraid it doesn't quite answer the question, since it doesn't use rot13.
$endgroup$
– Brandon_J
Mar 21 at 13:27
add a comment |
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$begingroup$
Don't have an answer, but I'll note 2 findings given the hint: 1) 505 is the hypotenuse of a triangle with integer length sides. 2) Rot13-ing the Roman Numerals for the number 105 produces a trigonometric term. Maybe one of these is on the right path and will help someone.
$endgroup$
– TwoBitOperation
Mar 20 at 14:40
3
$begingroup$
@TwoBitOperation for anyone possibly going with route 1: The possible sides of a triangle that work are (100, 495) (217, 456) (303, 404) (336, 377)
$endgroup$
– akozi
Mar 20 at 16:28